Concept: The question tests on the basic principles of counting/combinatorics

Solution:

In this question you have 2 numbers occurring the same number of times i.e twice and 1 number occurring once.

You can choose the one number occurring once in 9C1 = 9 ways and the 2 numbers occurring twice in 8C2 = 28 ways and then rearrange all these numbers in (5! / 2!2!) = 30 ways. Total number of ways= 9 * 28 * 30 = 7560

You can choose the two numbers occurring twice in 9C2 = 36 ways and the one number occurring once in 7C1 = 7 ways and then rearrange all these numbers in (5! / 2!2!) = 30 ways. Total number of ways= 36 * 7 * 30 = 7560

You can also choose on number which will occur twice in 9C1 = 9 ways, the next number occurring twice in 8C1 ways and the number occurring once in 7C1 ways.

What we need to remember here is that 2 of these numbers are occurring twice and hence we have to divide by 2! = 2 ways to avoid double counting.

Total number of ways = (9 * 8 * 7/2) * 30 = 7560 ways



 Option E